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Suppose that $f(z)$ is holomorphic and satisfies the condition $|f(z)^2-1|<1$ in a region $\Omega$. Show that either $\Re f(z)>0$ or $\Re f(z)<0$ throughout $\Omega$. ($\Re$ denotes the real part.)

I can factor $|f(z)-1|\cdot|f(z)+1|<1$, or if we write $f(z)=u(z)+iv(z)$, we have $$|(u(z)-1)+iv(z)|\cdot|(u(z)+1)+iv(z)|<1.$$

The goal is to prove that either $u(z)<0$ or $u(z)>0$ throughout the region, so perhaps a start is to assume that $u(z_1)<0$ and $u(z_2)>0$. Substituing in, we have $$|(u(z_1)-1)+iv(z_1)|\cdot|(u(z_1)+1)+iv(z_1)|<1.$$ Clearly the first absolute value term is greater than $1$, so $|(u(z_1)+1)+iv(z_1)|<1$, and similarly $|(u(z_2)-1)+iv(z_2)|<1$.

Also, there is a theorem that a holomorphic function in a region $\Omega$ whose derivative vanishes identically must reduce to a constant, but it doesn't seem to help.

Mika H.
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  • What do the conditions in the problem tell you about the distance of $f$ from $0$? Slightly more generally, what do they tell you about the distance from the imaginary axis? – Connor Aug 29 '13 at 16:48
  • I don't think it tells much... writing $f=u+iv$, we get $|(u^2-v^2-1)+(2uv)i|<1$, so we know $|u^2-v^2-1|<1$ and $|uv|<1/2$. How does that help? – Mika H. Aug 29 '13 at 17:10
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    Can $f(z)$ be purely imaginary (real part 0) given the conditions? My immediate reaction is to use this and invoke continuity. – Evan Aug 29 '13 at 17:16
  • As per Connor's comment, couldn't we start by saying that f^2(z) is entirely contained in the disk of radius 1 centered a (1,0)? Then so is f(z) which means its real part is > 0. The other solution -f(z) is contained in the disk of radius 1 centered at (-1,0), so it's real part is negative. Does this work? It seems simpler than factoring. – Betty Mock Aug 29 '13 at 17:49
  • @BettyMock Your step from $f^2$ to $f$ isn't immediately obvious (while it is true). You would have to say something like the square root for a point in $B_1(1)$ reduces the angle to the real axis while pushing the magnitude towards $1$, so the image is strictly contained in $B_1(1)$. But either way works! Note you still have to use continuity to show that all of $f$ is on one side or the other. – Evan Aug 29 '13 at 18:06
  • @Evan, I'm just a little confused. If f(z) is contained in the unit disk centered at (1,0) how can Re f not be all positive? I don't see what continuity has to do with it, let alone being holomorphic. Am I missing something? – Betty Mock Aug 30 '13 at 22:01
  • @BettyMock For each $f(z)^2$, either $f(z)$ is in the disk around $-1$ or the disk around $1$, as you say, but how do you rule out some values of $f(z)$ being in the $-1$ camp and other values of $f(z)$ being in the $1$ camp? Continuity shows that they all have to be on the same side. – Evan Aug 30 '13 at 22:14

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Ok, first we'll factor as you suggested:

$$|f(z)^2-1|=|f(z)+1||f(z)-1|$$

This tells us that there is a point $z$ where $|f(z)-1|<1$ or $|f(z)+1|<1$, i.e., the distance from $f(z)$ to $1$ is less than $1$ or the distance from $f(z)$ to $-1$ is less than one, respectively. Assume the former is true, so $\Re f(z)>0$. Now suppose there is another point $w$ with $\Re f(w)<0$. By continuity, there is a point $y$ with $\Re f(y)=0$. But if this is true, both $|f(y)-1|\ge 1$ and $|f(y)+1|\ge 1$, a contradiction.

Connor
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  • Hmm... so we don't need to do the analysis in your first 3-4 lines, do we? We can go straight to assuming $\Re f(z)>0$ and $\Re f(w)<0$. – Mika H. Aug 29 '13 at 17:30
  • @MikaH. was trying to add a bit of a geometric interpretation; it shows additionally that $f(\Omega)$ has to be contained entirely in a ball centered at $1$ or $-1$. I apologize if you'd have preferred a one-line answer. – Connor Aug 29 '13 at 17:33
  • Certainly no need to apologize, Connor. I just wanted to make sure that I didn't misunderstand something as I couldn't see the necessity of the first part. Thanks! – Mika H. Aug 29 '13 at 17:37